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- Thread starter mt91
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- Jan 30, 2018

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So $\frac{du*}{dt}= 0= u*(1- u*)(1+ u*)- Eu*$.

Solve that third degree equation for u*.

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Cheers, it was the wording of the question I struggled with, because it said as a function of E. So this is what I did:constantsolution to the differential equation. And if it is constant, $\frac{du*}{dt}= 0$

So $\frac{du*}{dt}= 0= u*(1- u*)(1+ u*)- Eu*$.

Solve that third degree equation for u*.

\[ Eu*=u*(1-u*)(1+u*) \]

\[ E=(1-u*)(1+u*) \]

\[ ∴u*=1-E, u*=E-1 \]

I think that u*=0 would be a solution, but I'm not sure with the wording of "function of E" what it wants me to do, cheers

- Jan 30, 2018

- 464

Yes, Eu*= u*(1- u*)(1+ u*). u*= 0 is a solution. If u* is not 0 we can divide by u* to get E= (1- u*)(1+ u*).

But that does NOT give u*= 1- E or u*= 1+ E as solutions! You would have seen that if you had checked: If u*= 1- E then 1- u*= 1-1+ E= E and 1+ E= 2- E. $(1- u*)(1+ u*)= E(2- E)= 2E- E^2$.

From $E= (1- u*)(1+ u*)= 1- u*^2$ we get $u*^2= 1- E$ so that $u*= \pm\sqrt{1- E}$.

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Meaning therefore we have 3 steady state solutions?

Yes, Eu*= u*(1- u*)(1+ u*). u*= 0 is a solution. If u* is not 0 we can divide by u* to get E= (1- u*)(1+ u*).

But that does NOT give u*= 1- E or u*= 1+ E as solutions! You would have seen that if you had checked: If u*= 1- E then 1- u*= 1-1+ E= E and 1+ E= 2- E. $(1- u*)(1+ u*)= E(2- E)= 2E- E^2$.

From $E= (1- u*)(1+ u*)= 1- u*^2$ we get $u*^2= 1- E$ so that $u*= \pm\sqrt{1- E}$.

\[ u*=0 \]

\[ u*=sqrt(1-E) \]

\[ u*=-sqrt(1-E) \]

Thanks for your help by the way

- Jan 30, 2018

- 464

Yes.

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Thanks, do you know much about how you would interpret relevant steady states? So the steady state equation was about an animal population. If a steady state is 0 is this indicating that there's no population? and the negative square root steady state is indicating a decline in the population?Yes.

- Jan 30, 2018

- 464

The population is u(t). If for any t, u(t)= 0, there is no population so, of course, it cannot decrease and is not going to increase so stays 0. 0 is always a "steady state" for a population. (Unless you allow "spontaneous generation"!)Thanks, do you know much about how you would interpret relevant steady states? So the steady state equation was about an animal population. If a steady state is 0 is this indicating that there's no population?

No, the whole point of "steady state" is that it isand the negative square root steady state is indicating a decline in the population?